Mathematics 9 Quarter 2 Module 9: Solving Equations Involving Radical Expressions

Mathematics Quarter 2 – Module 9 Solving Equations Involving Radical Expressions

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Mathematics – Grade 9 Alternative Delivery Mode Quarter 2 – Module 9 :Solving Equations Involving Radical Expressions First Edition, 2020

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Mathematics Quarter 2 – Module 9: Solving Equation Involving Radical Expressions

Introductory Message

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

Thank you.

Lesson 1 Radical Equations

In the previous lessons, you have learned how to perform operations on radical expressions. This time, you will learn how to apply these concepts in solving radical equations.

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What I Need to Know

LEARNING COMPETENCY:

The learners will be able to:

Solve equations involving radical expressions. M9AL – IIi – j – 27

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What I Know

Directions: Choose the letter of the correct answer.

1. Solve for x in the equation √(x − 3) = 1.
   a. x = 4
   b. x = −4
   c. x = 2
   d. There are no real solutions

   
   

2. Solve the radical equation: √(3x) = 12
   a. 12
   b. 48
   c. 24
   d. 36

   
   

3. Which of the following are value(s) of x that will satisfy the equation √(5x + 16) − x = 2?
   a. 3 and 14
   b. 4
   c. −1 and 5
   d. −3

   
   

4. Solve the equation: √(3x − 2) = 6.
   a. 4
   b. 14/3
   c. 38/3
   d. 9

   
   

For numbers 5-8, refers to the choices below: a. x = 1 b. x = 33 c. x = 2 d. x = 20

What is the solution of the equation?

5. √(3x + 1) − 3 = 7

   
   

6. √(19 − 3x) − 1 = 3x

   
   

7. √(2x − 15) = √(x + 5)

   
   

8. √(x + 2) + √(x − 1) = 3

   
   

9. If the given value of a solution is n = 10, what is the equation?
   a. √(110 − n) = n
   b. √(30 − x) = x
   c. x = √(8x)
   d. √(4n) = n

   
   

10. Which of the following value(s) of x satisfy the equation √(6x + 4) = 4?
    a. x = 2
    b. x = 4
    c. x = 3
    d. no solution

    
    

11. Solve the equation: √(x − 1) = 3
    a. x = −10
    b. x = 3
    c. x = −3
    d. x = 1

    
    

For numbers 12-15, refers to the choices below: a. x = 9 b. x = 21 c. no solution d. x = 5, x = 7

Find the solution of the equation.

12. −6 + √(x − 5) = −2

    
    

13. √(x + 72) = x

    
    

14. 8√(x + 6) − 7 = −79

    
    

15. x = 5 + √(2x − 10)

    
    

What’s In

The following operations on radical expressions will be needed in solving equations involving radical equations:

1. Addition and Subtraction: Write all radicals in their simplest form then combine like radicals. (Like radicals have the same indices and radicand).
2. Multiplication: Apply the product rule for radicals (ⁿ√(ab) = ⁿ√a · ⁿ√b), then simplify. Apply distributive property when multiplying radicals with two or more terms.
3. Division: Write the quotient as fraction form and rationalize the denominator. We may also apply the quotient rule for radicals (ⁿ√(a/b) = ⁿ√a / ⁿ√b), then simplify.

Let us review these operations by answering the exercises below.

Perform the indicated operations:

1. 8√5 + √5
2. 5√3 − 3√3
3. √8 + √20 − √12
4. 3√20 − 2√45
5. ∛(2a) · √(2a)
6. √50 · √(2m⁵)
7. ⁴√14 ÷ ⁴√2
8. √(21x⁷) ÷ √(3x²)
9. ⁴√a (⁴√(a³) − 7)
10. 7 ÷ (√6 + √5)

Solve the following:

11. Maggie walks around a rectangular field. The field has a length of √27 meters and is 2√3-meter wide. What is the total distance Maggie walks through the field?

    
    

12. The base of an isosceles triangle measures 10√2 units while its height measures 12√2 units. Express the area of the triangle as an exact value in simplest form.

    
    

What’s New

Mathematical problems that involve radical equations arise in a number of professions. We have rational equations that are used to calculate depreciation, inflation and interest rates. Engineers used radical equations for their measurements and calculations. Scientists use radical exponents for size comparison in research. Let us take a look at how these equations are solve

[chart: flowchart; ACTIVITY 1 SOLVING RADICAL EQUATIONS – STUDENT’S WORKSHEET ON EXAMPLE – CONCLUSION MAP (ECM)]

Example 1 Solve for y: √y = 15

Solution	Reason
√y = 15	Given
(√y)² = 15²	Principle of Powers
y = 225	Simplify
Answer: 225

Example 2 Solve for x: √(x − 3) = 1

Solution	Reason
√(x − 3) = 1	Given
√x = 1 + 3	Isolate the radical using APE
(√x)² = 4²	Principle of Powers
x = 16	Simplify
Answer: 16

Example 3 Solve for x: √(2x − 3) − 5 = 0

Solution	Reason
√(2x − 3) − 5 = 0	Given
√(2x − 3) = 5	Isolate the radical
(√(2x − 3))² = 5²	Principle of Powers
2x − 3 = 25
2x = 28	APE
x = 14	MPE
Answer: 14

Conclusion

To solve a radical equation, follow these steps. (1) _____ the radical, make sure that one radical term is alone on one side of the equation. (2) Apply the _____, raise each side of the equation to a power that is the same as the index of the radical. (3) _____. Solve the resulting equation. If it still contains radical, repeat step 1 and step 2.

Your own example

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Solving Radical Equations

A radical equation is an equation that has variables in one or more radicands. Examples are √x = 8 and 2 = ∛m. To solve these kinds of radical equations, we need to apply the Principle of Powers. That is, if a = b, then aⁿ = bⁿ.

Usually, radical equations are not as simple as the examples given above. Most of the time, the equation contains several terms on each side. To solve equations of this kind, the following steps must be followed.

1. Isolate the radical.
2. Apply the principle of powers.
3. Solve the equation. If there is still a radical, repeat 1 and 2.
4. Check for extraneous solution.

Examples

1. Solve √x = 8.

Solution: √x = 8
          (√x)² = 8²                  Principle of Powers
          x = 64
Check:    √x = 8
          √64 = 8                     Substitution
          8 = 8                       Therefore, the solution/value of the
                                      variable is acceptable

2. Solve 2 = ∛m.

Solution: 2 = ∛m
          2³ = (∛m)³                  Principle of Powers
          8 = m
Check:    2 = ∛m
          2 = ∛8                      Substitution
          2 = 2                       Therefore, the solution/value of the
                                      variable is acceptable

3. Solve √(x + 2) = 3

Solution: √(x + 2) = 3
          (√(x + 2))² = 3²            Principle of Powers
          x + 2 = 9
          x = 9 − 2                   Addition Property of Equality
          x = 7                       Simplify
Check:    √(x + 2) = 3
          √(7 + 2) = 3                Substitution
          √9 = 3
          3 = 3                       Therefore, the solution/value of the
                                      variable is acceptable

4. Solve √(3x + 2) = √(x + 10)

Solution: √(3x + 2) = √(x + 10)
          (√(3x + 2))² = (√(x + 10))²  Principle of Powers
          3x + 2 = x + 10
          3x − x = 10 − 2             APE
          2x = 8
          2x/2 = 8/2                  MPE
          x = 4                       Simplify
Check:    √(3x + 2) = √(x + 10)
          √(3 · 4 + 2) = √(4 + 10)    Substitution
          √14 = √14                   Therefore, the solution/value of the
                                      variable is acceptable

5. Solve ∛(x + 3) = 2.

Solution: ∛(x + 3) = 2
          (∛(x + 3))³ = 2³            Principle of Powers
          x + 3 = 8
          x = 8 − 3                   APE
          x = 5                       Simplify
Check:    ∛(x + 3) = 2
          ∛(5 + 3) = 2                Substitution
          ∛8 = 2
          2 = 2                       Therefore, the solution/value of the
                                      variable is acceptable

6. Solve √(5a + 1) + 2 = 6

Solution: √(5a + 1) + 2 = 6
          √(5a + 1) = 6 − 2           APE
          (√(5a + 1))² = 4²           Principle of Powers
          5a + 1 = 16
          5a = 16 − 1                 APE
          5a = 15
          5a/5 = 15/5                 MPE
          a = 3                       Simplify
Check:    √(5a + 1) + 2 = 6
          √(5 · 3 + 1) + 2 = 6        Substitution
          √16 + 2 = 6
          4 + 2 = 6
          6 = 6                       Therefore, the solution/value of the
                                      variable is acceptable

7. Solve √(7x + 2) − 2x = 0

Solution: √(7x + 2) − 2x = 0
          √(7x + 2) = 2x              APE
          (√(7x + 2))² = (2x)²        Principle of Powers
          7x + 2 = 4x²
          0 = 4x² − 7x − 2            APE
          4x² − 7x − 2 = 0            Symmetric Property of Equality
          (4x + 1)(x − 2) = 0         Factoring
          4x + 1 = 0 ; x − 2 = 0      Equate both factors to 0
          4x = −1 ; x = 2             APE
          x = −1/4 ; x = 2            MPE

Check: By substitution

x = −1/4                              x = 2
√(7x + 2) − 2x = 0                    √(7x + 2) − 2x = 0
√(7(−1/4) + 2) − 2(−1/4) = 0          √(7 · 2 + 2) − 2 · 2 = 0
√(−7/4 + 2 + 2/4) = 0                 √(14 + 2) − 4 = 0
√((−7 + 8)/4 + 1/2) = 0               √16 − 4 = 0
√(1/4 + 1/2) = 0                      4 − 4 = 0

√(1/4 + 1/2) = 0 and 0 = 0
1/2 + 1/2 = 0 and 0 = 0
1 ≠ 0

One value (−1/4) of x did not satisfy the equation. Therefore, it is not a solution. The solution of the given radical equation is just 2.

What’s More

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[chart: flowchart; ACTIVITY 2 Solving Radical Equations, Student’s Worksheet on Math-Breaker Map (MBM Structured)]

Solve the radical equation: ∛(6x + 12) + 4 = 12

Boxes, in order: ISOLATE THE RADICAL → APPLY PRINCIPLE OF POWERS → SOLVE THE LINEAR EQUATION → WRITE THE FINAL ANSWER (the boxes beneath the first two labels are empty)

A. Read each item carefully and determine whether the statement is true or false.

1. √5 + x = x² is an example of a radical equation.
2. Radical equations may not always have real solutions.
3. The value x = -4 is a solution to √(5 − x) = x + 7.
4. The √(x + 1) = 2 has two solutions.
5. The ∛(3x + 1) = −2 has a positive solution.

B. Find the value(s) of x in the following radical equations:

1. √x − 3 = 5
2. √(x + 8) = 3
3. ∛(2x + 3) + 5 = 0
4. √(3x − 1) − √(x + 5) = 0
5. x = √(x − 4) + 4
6. x + 4 = √(x + 10)

What I Have Learned

The solution of a radical equation is a value that will satisfy the given equation. To solve a radical equation, we follow these steps:

1. Isolate the radical on one side of the equation.
2. Apply the Principle of Powers.
3. If radical remains, repeat steps 1 and 2.
4. Check if the obtained value or values satisfy the given equation.

Critical Thinking

A. Analyze each item and identify on which part, the solution started to get wrong. Write down the letter only.

1. √(20 − n) = n

(√(20 − n))² = n²                     (A)
20 − n = n²                           (B)
0 = n² − n − 20                       (C)
n² − n − 20 = 0
(n − 5)(n + 4) = 0
n = 5 or n = −4

2. √x = −3

(√x)² = (−3)²                         (A)
x = 9                                 (B)
The solution to the radical equation is 9.  (C)

3. √(x + 3) − 1 = 7

√(x + 3) = 7 − 1                      (A)
√(x + 3) = 6
(√(x + 3))² = 6²                      (B)
x + 3 = 36
x = 33                                (C)

4. √x = √(x + 2)

(√x)² = (√(x + 2))²                   (A)
x² = x + 2                            (B)
x² − x − 2 = 0                        (C)
(x − 2)(x + 1) = 0
x = 2 or x = −1

5. √(2x)³ … [transcriber’s note: the radical in item 5 is printed as a cube root, ∛(2x) = −4]

∛(2x) = −4
(∛(2x))³ = (−4)³                      (A)
√(2x) = −4                            (B)
(√(2x))² = (−4)²
2x = 16
x = 4                                 (C)

B. Solve for the value(s) of x.

1. √(2x) = 6 6. ∛x − 1 = 2
2. 3√(4x) = 36 7. √(25 − x²) = 4
3. √(x − 7) = 3 8. 1 + √(2x + 3) = 6
4. √x + 13 = 20 9. ∛(x² + 28) = 4
5. √(5x + 11) − 1 = x 10. √(2x − 2) = x − 1

What I Can Do

PROBLEM – BASED LEARNING WORKSHEET

HERON’S FORMULA

The area (A) of a triangle is given by the formula A = ½bh, where b is the length of the base and h is the height. However, we can still compute for the area of a triangle using the lengths of its three sides. The Greek mathematician Heron proved this by developing a formula that uses the semiperimeter (½ the perimeter) of the given triangle.

If a triangle has sides whose lengths are a, b and c and s is ½ the sum of a, b and c, the area (A) of the triangle is given by

A = √(s(s − a)(s − b)(s − c))

LET’S ANALYZE

1. Who developed the formula?
2. What do you mean by semiperimeter?
3. Give the two formulas used to find the area of a triangle.
4. Analyn has a triangular lot whose dimensions are shown below. What is the lot’s perimeter? Semiperimeter?

[chart: diagram; triangular lot of Analyn, sides and height labeled as printed] Labels: 5 m (left side), 5 m (right side), 4 m (height, shown as a perpendicular from the apex to the base), 6 m (base); a right-angle mark is shown at the foot of the height.

5. Using the Heron’s formula, compute for the area of Analyn’s lot. Check your answer using the other formula.
6. An equilateral triangle has an area of 9√3 units. How long is each side of the triangle?

Assessment

Read the questions carefully. Write the letter of the correct answer on your answer sheet.

1. A ________ is an equation with one unknown in one or more of the radicals.
   a. quadratic equation
   b. linear equation
   c. radical equation
   d. none of these

   
   

2. What value(s) of y satisfy the equation √(5y + 1) = √(3y + 1) ?
   a. 0
   b. 9/4
   c. 0, 9/4
   d. no solution

   
   

3. The value of x in the equation ∛(x + 1) = 3
   a. 23 b. 26 c. 27 d. 29

   
   

(4-5 Refer to the given problem below) The square root of the two consecutive odd integers is equal to 6.

4. What is the equation in terms of y?
   a. √y + √(y + 2) = 6
   b. √(y + (y + 2)) = 6
   c. √(y + √(y + 2)) = 6
   d. √(y + (y + 2)) = √6

   
   

5. Find the larger integer.
   a. 17 b. 10 c. 6 d. 19

   
   

(6-8 Solve and check)

6. 2√(x − 3) = 12
   a. 36 b. 30 c. 39 d. 35

   
   

7. √(5x + 4) + 6 = 4
   a. 0 b. 2 c. 4 d. no solution

   
   

8. √(x − 2) = √(3x − 14)
   a. 2 b. 4 c. 6 d. no solution

   
   

9. What is the solution of √(6x − 4) = 3
   a. 4/3 b. −7/6 c. 13/6 d. −13/6

   
   

10. The square root of the sum of two consecutive integers is 7. Find the two integers.
    a. 3 and 4
    b. 11 and 12
    c. 17 and 18
    d. 24 and 25

    
    

Additional Activities

Critical Thinking & Collaboration

Who is this Mathematician?

This Welsh Mathematician introduced the equal sign in 1557.

To find out:

1. Find the solution to each radical equation.
2. Write the letter that corresponds to the solution in the decoder below.

[transcriber’s note: the original layout is a grid of seven cells in two rows; it is split here into two tables.]

B	C	D	E
√x = 7	12 = √(a + 108)	√(2y + 9) = 1	√(5n + 9) + 4 = 7

O	R	T
√(10 − 3x) = √(2x + 20)	∜(x + 8) = ∜(3x)	4 = √(x + 7)

DECODER:

_____ 4 _____ -2 _____ 49 _____ 0 _____ 4 _____ 9

_____ 4 _____ 0 _____ 36 _____ -2 _____ 4 _____ -4 _____ 0

E-Search

To further explore the concept learned today and if it’s possible to connect the internet, you may visit the following links:

https://www.youtube.com/watch?v=0gicD4STzpg https://www.youtube.com/watch?v=w79rpKCKlFw

Answer Key

[transcriber’s note: this page is printed upside down; the text below is in reading order, with the page’s left column first]

WHAT I KNOW

1. A
2. B
3. B
4. C
5. B
6. A
7. D
8. C
9. A
10. A
11. D
12. B
13. A
14. C
15. D

WHAT’S IN

1. 9√5
2. 2√3
3. 2√2 + 2√5 − 2√3
4. 0
5. √(2s²)/a⁵ [?]
6. 10m²√(vm) [?]
7. √7/√7 [?]
8. x²√(7x)/√(7x) [?]
9. a − 7√a [?]
10. 7(√6 − √5) [?]
11. 10√5 yards
12. 120√2 units [?]

WHAT’S MORE

A.

1. False
2. True
3. True
4. False
5. False

B.

1. x = 64
2. x = 1
3. x = -64
4. x = 3
5. x = 5, 4
6. x = -1

WHAT I CAN DO

A.

1. C, it should be 0 = n² + n − 20
2. C, there is no real solution.
3. A, it should be √(x + 3) = 7 + 1
4. B, it should be x = x + 2
5. B, it should only be 2x = -64

B.

1. x = 18
2. x = 36 [?]
3. x = 16
4. x = 49
5. x = 5
6. x = -27
7. x = ±43 [?]
8. x = ±11 [?]
9. x = ±6 [?]
10. x = -1, 3

ASSESSMENT

1. C
2. A
3. B
4. B
5. D
6. C
7. D
8. C
9. C
10. D

ADDITIONAL ACTIVITIES

ROBERT RECORDE

PROBLEM – BASED LEARNING WORKSHEET

1. Heron, a Greek Mathematician
2. Half of the perimeter
3. A = √(s(s − a)(s − b)(s − c)) and A = ½ bh
4. P = 16 m; s = 8 m; A = √(s(s − a)(s − b)(s − c))
5. A = √(s(s − a)(s − b)(s − c))
   A = √(8(8 − 5)(8 − 3)(8 − 6))
   A = √(8(3)(3)(2)) [?]
   A = 12 sq. m
   A = ½ bh
   A = ½ (6)(4)
   A = 12 sq. m
6. 6 units

References

[chart: concept map; ACTIVITY 1 SOLVING RADICAL EQUATIONS STUDENT’S WORKSHEET ON EXAMPLE – CONCLUSION MAP (ECM)]

Example 1 Solve for y: ∛y = 15 Given: ∛y = 15 Solution: (∛y)³ = 15³ Principle of Powers; y = 15³ … Answer: 225 [?] Simplify

Example 2 Solve for x: √(x − 3) = 1 Given: √(x − 3) = 1 Solution: Isolate the radical using APE: √x = 1 + 3; (√x)² = 4² Principle of Powers; x = 16 Simplify Answer: 16

Example 3 Solve for x: √(2x − 3) − 5 = 0 Given: √(2x − 3) − 5 = 0 Solution: Isolate the radical: √(2x − 3) = 5; (√(2x − 3))² = 5² Principle of Powers; 2x − 3 = 25 Simplify; 2x = 28 APE; x = 14 MPE Answer: 14

Conclusion: To solve a radical equation, follow these three steps: (1) Isolate the radical, make sure only one radical term is on one side of the equation. (2) Apply the Principle of Powers: raise each side of the equation to the power equal to the index of the radical. (3) Simplify. Solve the resulting equation. If it still contains a radical, repeat steps 1 and 2.

Your own example: Answers may vary.

[chart: concept map; ACTIVITY 2 SOLVING RADICAL EQUATIONS STUDENT’S WORKSHEET ON MATH-BREAKER MAP (MBM STRUCTURED)]

Solve the radical equation: ∛(6x + 12) + 4 = 12 Isolate the radical: ∛(6x + 12) = 12 − 4 Apply principle of powers: (∛(6x + 12))³ = 8³ Solve linear equation: 6x + 12 = 512 [?] 6x = 500 x = 250/3

[transcriber’s note: the flowchart arrows and box order are as printed; the exact printed text of some lines is unclear in the rotated image]

References

Nivera, G.C. & Lapinid, M. C. (2013). Grade 9 Mathematics Patterns and Practicalities. Salesiana Books by Don Bosco Press

Oronce, O. A. & Mendoza, M. O. C. (2003). Exploring Mathematics II Intermediate Algebra, Rex Book Store DepEd Project Ease Mathematics Module 6 (downloaded from LRMDS) Validated Problem Solving Maps Worksheets

Prentice Hall Algebra 1 Teaching Resource https://1.cdn.edl.io/x1HxFjyVLo8pgAr2evQRFJb68knnbUOXdiUn0W67oRsODBSt. pdfhttp://jeff560.tripod.com/mathword.html

https://tl.wikipedia.org/wiki/Robert_Recorde

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